Problem 135 - Calculus 3

Question

For the following problems, find the solution to the boundary-value problem. $$4 y^{\prime}=-6 y+2 y^{\prime \prime}, \quad y(0)=0, \quad y(1)=1$$

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{\sin(\sqrt{2}x)}{\sin(\sqrt{2})} \).

Step 1: Rearrange the differential equation

The given differential equation is \( 2y^{\prime \prime} - 4y^{\prime} = -6y \). To work with it in a more standard form, divide every term by 2 to get: \( y^{\prime \prime} - 2y^{\prime} = -3y \). This is a second-order linear differential equation.

Step 2: Solve the characteristic equation

To find the solution to the homogeneous equation, we use its characteristic equation. For \( y^{\prime \prime} - 2y^{\prime} + 3y = 0 \), the characteristic equation is \( r^2 - 2r + 3 = 0 \). Solving this quadratic equation using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), gives: \( r = 1 \pm i \sqrt{2} \).

Step 3: Write the general solution of the homogeneous equation

The roots \( r = 1 \pm i \sqrt{2} \) imply complex conjugates. The general solution for complex roots \( \alpha \pm i \beta \) is \(y_h(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x))\). Therefore, the homogeneous solution is \( y_h(x) = e^x(C_1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x)) \).

Step 4: Apply boundary conditions

Apply the boundary conditions \( y(0)=0 \) and \( y(1)=1 \) to find \( C_1 \) and \( C_2 \). At \( x = 0 \), the equation becomes \( y_h(0) = C_1 = 0 \) (since \( e^0 = 1 \) and \( \sin(0) = 0 \)). Applying \( y(1) = 1 \) gives \( e(0 \cdot \cos(\sqrt{2})) + C_2 \cdot \sin(\sqrt{2}) = 1 \), simplifying to \( C_2 = \frac{1}{e \sin(\sqrt{2})} \).

Step 5: Write the particular solution using C1 and C2

Substituting \( C_1 = 0 \) and \( C_2 = \frac{1}{e \sin(\sqrt{2})} \) into the homogeneous solution gives \( y(x) = e^x \frac{1}{e \sin(\sqrt{2})} \sin(\sqrt{2}x) \), which simplifies to \( y(x) = \frac{\sin(\sqrt{2}x)}{\sin(\sqrt{2})} \). This satisfies both the differential equation and the boundary conditions.

Key Concepts

Second-Order Linear Differential Equations

A second-order linear differential equation is a key concept in calculus and differential equations, typically represented in the form:
\[ a x^2 y'' + b x y' + c y = f(x) \]It involves the function \( y \) and its derivatives \( y' \) and \( y'' \). In our given problem, after simplification, the standard form is:
\[ y'' - 2y' = -3y \]Here,:

• "\( y'' \)" represents the second derivative, indicating how \( y \) changes at changing rates.
• "\( -2y' \)" modifies the rate of change.
• "\( -3y \)" defines how our function itself influences these changes.

This type of differential equation is crucial because it describes various real-world phenomena, such as oscillations or growth rates, helping us model scenarios like sound waves or population dynamics.
Understanding these equations facilitates solving boundary-value problems, as they define the dynamics of the entire system. Recognizing and organizing these equations correctly is a fundamental step in mathematical modeling and problem-solving.

Characteristic Equation

The characteristic equation is derived from a second-order homogeneous linear differential equation. Its central role is to identify the nature of solutions by transforming the differential equation into an algebraic form. For our problem, the differential equation:
\[ y'' - 2y' + 3y = 0 \]transforms into its characteristic equation:
\[ r^2 - 2r + 3 = 0 \]By solving this quadratic equation, we determine the roots, which show how solutions behave. The characteristic equation uses the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substituting the coefficients gives:
\[ r = 1 \pm i\sqrt{2} \]These roots are complex, indicating oscillatory solutions typical in problems involving physical systems and engineering contexts. This insight helps us form a framework for the general solution, which includes exponential and trigonometric components.

Complex Roots

Having complex roots \( r = \alpha \pm \beta i \) often signals solutions involving oscillations. These roots are crucial in forming the general solution of the differential equation. For roots like \( r = 1 \pm i \sqrt{2} \), the general solution takes the form:
\[ y_h(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \]This formula results from Euler's formula, converting complex exponents into sine and cosine functions. Here's how it unfolds for our example:

• \( e^{\alpha x} = e^x \), provides the exponential growth or decay.
• \( C_1 \cos(\beta x) + C_2 \sin(\beta x) \), where \( \beta = \sqrt{2} \), indicates oscillatory components.

These components represent the real and imaginary parts of complex solutions, bringing the oscillations of waveforms, alternating currents, and vibrations into play. Identifying such solutions is vital in systems that involve periodic behavior, improving our understanding and prediction of natural and engineered phenomena.

Boundary Conditions

Boundary conditions are specific requirements the solution must satisfy, given at particular points in the domain, such as ends of an interval. They are critical to finding unique solutions to differential equations.
In our case:

• \( y(0) = 0 \)
• \( y(1) = 1 \)

These set conditions allow us to determine constants in our general solution. Without boundary conditions, we'd have infinite possible solutions. Applying these conditions:

• At \( x = 0 \), we find \( C_1 = 0 \) from \( y_h(0) = C_1 \)
• At \( x = 1 \), using \( y(1) = 1 \), we solve for \( C_2 \)

The solution then becomes specific to our problem, integrating the mathematical model with practical constraints. Mastering boundary conditions in problem-solving helps craft accurately tailored solutions in science and engineering fields, crucial for creating effective designs and systems.